//Sylwia Sapkowska
#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
using namespace std;
void __print(int x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << "'" << x << "'";}
void __print(const char *x) {cerr << '"' << x << '"';}
void __print(const string &x) {cerr << '"' << x << '"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifdef LOCAL
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif
#define int long long
typedef pair<int, int> T;
const int oo = 1e18, oo2 = 1e9+7, K = 30;
const int mod = 998244353;
void solve(){
int n, m; cin >> n >> m;
vector a(n+1, vector<int>(m+1));
int mx = -oo, mn = oo;
T A, B;
for (int i = 1; i<=n; i++){
for (int j = 1; j<=m; j++){
cin >> a[i][j];
if (a[i][j] > mx){
mx = a[i][j];
A = {i, j};
}
if (a[i][j] < mn){
mn = a[i][j];
B = {i, j};
}
}
}
if (mn == mx){
cout << "0\n";
return;
}
int ans = mx-mn;
auto sortuj = [&](vector<int>L, vector<int>R){
debug(L, R); //a[i] \in [L[i], R[i]]
//1) increasing
bool ok = 1;
vector<int>tab(n+1);
for (int i = 1; i<=n; i++){
tab[i] = L[i];
if (i > 1) tab[i] = max(tab[i], tab[i-1]);
if (tab[i] > R[i] || tab[i] < L[i]){
ok = 0;
break;
}
}
if (ok) return true;
ok = 1;
//2) decreasing
for (int i = 1; i<=n; i++){
tab[i] = R[i];
if (i > 1) tab[i] = min(tab[i], tab[i-1]);
if (tab[i] > R[i] || tab[i] < L[i]){
ok = 0;
break;
}
}
return ok;
};
auto check = [&](int k){
//max(A-mn, mx-B) = k;
//A <= k + mn
//B >= mx - k
debug(k+mn, mx-k);
//sprawdzic czy da sie przypisac
//0) z lewa minimum
//1) z lewa maksimum
for (int rep = 0; rep < 2; rep++){
vector<int>R(n+1, m), L(n+1, 0);
bool ok = 1;
for (int i = 1; i<=n; i++){
for (int j = 1; j<=m; j++){
if ((!rep && a[i][j] > k + mn) || (rep && a[i][j] < mx - k)){
R[i] = j-1;
break;
}
}
for (int j = R[i]+1; j<=m; j++){
if (!rep) ok &= (a[i][j] >= mx-k);
else ok &= (a[i][j] <= k+mn);
}
for (int j = R[i]; j>=1; j--){
if ((!rep && a[i][j] < mx-k) || (rep && a[i][j] > k+mn)){
L[i] = j;
break;
}
}
}
if (ok && sortuj(L, R)) return true;
}
return false;
};
//ans = max(mx-A, B-mn);
int l = 1, r = oo;
while (r >= l){
int mid = (l+r)/2;
if (check(mid)){
ans = mid;
r = mid-1;
} else {
l = mid+1;
}
}
ans = min(ans, mx-mn);
cout << ans << "\n";
}
int32_t main(){
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t = 1;
//cin >> t;
while (t--) solve();
return 0;
}
